# Number of integral solutions of $\text{xyz}=3000$

I want to find the number of integral solutions of the equation $$xyz=3000$$

I have been able to solve similar sums where the number on the right hand side was small enough to calculate all the factors of. Such as $xyz=24$ or $xyz=30$.

What is the proper method to solve such a problem when the number is too big to consider all the factors of it?

• The solutions you have assume that $3 \cdot 8 \cdot 125$ is different from $125 \cdot 3 \cdot 8$. Is that what you intended? It is not clear from the question. May 1, 2015 at 3:55
• Yes. I want the solutions as ordered pairs of {x,y,z}. So, {3,8,125} will be different from {125,3,8}. May 6, 2015 at 2:31

$\bf{My\; Solution::}$ We Can write $x\cdot y\cdot z= 3000 = 2^3\cdot 3^1\cdot 5^3$

Now let $x=2^{x_{1}}\cdot 3^{y_{1}}\cdot 5^{z_{1}}$ and $2^{x_{2}}\cdot 3^{y_{2}}\cdot 5^{z_{2}}$ and $2^{x_{3}}\cdot 3^{y_{3}}\cdot 5^{z_{3}}$

So $x\cdot y \cdot z = 2^{x_{1}+x_{2}+x_{3}}\cdot 3^{y_{1}+y_{2}+y_{3}}\cdot 2^{z_{1}+z_{2}+z_{3}}=2^{3}\cdot 3^{1}\cdot 5^{3}$

So we get $x_{1}+x_{2}+x_{3}=3$ and $y_{1}+y_{2}+y_{3}=1$ and $z_{1}+z_{2}+z_{3} = 3\;,$

Where $0 \leq x_{1}+x_{2}+x_{3}\leq 3\;,0\leq y_{1}+y_{2}+y_{3}\leq 1\;,0\leq z_{1}+z_{2}+z_{3}\leq 3$

So we get $(x_{1}\;,x_{2}\;,x_{3}) = 10$ pairs. and $(y_{1},y_{2},y_{3}) = 3$ pairs and $(z_{1},z_{2},z_{3}) = 10$ pairs

So We Get $(x,y,z) = 10 \times 3 \times 10 = 300$ positive integer ordered pairs.

Now It is possible that any two variables is $(-)$ve and one is $(+)$ve.

So We get Total ordered pairs is $= \bf{all\; positive}+\bf{any\; two \; is \; negative.}$

• This is exactly what I wanted to understand! :) May 1, 2015 at 3:36

Note that $3000 = 2^3 \cdot 3^1 \cdot 5^3$. Let $x=2^{x_1}3^{x_2}5^{x_3}$, $y=2^{y_1}3^{y_2}5^{y_3}$ and $z=2^{z_1}3^{z_2}5^{z_3}$. Then obtain the number of solutions to $$x_1+x_2+x_3 = 3$$ $$y_1+y_2+y_3 = 1$$ $$z_1+z_2+z_3 = 3$$ The product of the number of non-negative solutions will give us the total number of solutions. Hence, in our case, the number of non-negative solutions to the first one is $\dbinom{3+3-1}{3-1} =10$, the second one is $\dbinom{1+3-1}{3-1} =3$ and the third one is $\dbinom{3+3-1}{3-1} =10$. Hence, total number of solutions is $10\cdot 3 \cdot 10 = 300$.

Hint: factorize $3000$ and do permutations of the factors.

• $3000$ has a huge number of factors. Isn't it a difficult task to find all the permutations? May 1, 2015 at 3:15
• @Tejas 3000 only has 7 (prime) factors. May 1, 2015 at 3:16
• $3000=2^3.3^1.5^3$. How $7$? May 1, 2015 at 3:18
• 2, 2, 2, 3, 5, 5 and 5. May 1, 2015 at 3:20
• Oh. Ok. But I'm still unclear about it. How do I proceed? May 1, 2015 at 3:22

$3000=3\cdot2^3\cdot5^3$.

First choose how to distribute the $3$'s, there are $3$ ways to do this.

Then choose how to distribute the $2$'s, there are $3+6+1=10$ ways to do this.

Finally distribute the $5$'s, there are $10$ ways to do this also.

Therefore there are $3\cdot 10 \cdot 10=300$ ways to split $3000$ into $3$ factors $x,y,x$ where the order matters.

You can solve the general problem with stars and bars.

In case of total number of integral solutions: Let $$x_1x_2...x_r = p_1^{a_1} p_2^{a_2} ...p_k^{a_k}$$ All the factors in each factorization has two options : $$(+)$$ or $$(-)$$. So, the result is $$\prod_{i=1}^k (2^{a_i} - \sum_{j=1}^l\binom{a_i}{2j-1}) × \prod_{i=1}^k \binom{a_i +r -1}{r-1}$$ where $$l = \frac{a_i}{2}$$ if $$a_i$$ is even and $$l= \frac{a_i+1}{2}$$ if $$a_i$$ is odd.